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%% Chapter 2, set the counter of section to i-1 for chapter i
\setsection{1}

\section{Normed Linear Spaces \& Banach Spaces}

\begin{Exec}
  Suppose that $(X,\|\cdot\|)$ is normed linear space.
  If $x\in X\backslash\{0\}$ and $r>0$, find a real
  number $c\in\br$ such that $\|c x\|=r$.
\end{Exec}
\begin{proof}
  Since $x\in X\setminus\{0\}$, $\|x\| \neq 0$.
  Let $c = r/\|x\|$, then $\|cx\| = c \|x\| = r$.
\end{proof}

\begin{Exec}\label{Exec2.2}
  Suppose that $(X,\|\cdot\|_X)$ and $(Y,\|\cdot\|_Y)$
  are normed linear spaces over $\bbf$.
  Consider the cartesian product $X\times Y$ with vector addition and
  scalar multiplication defined by
  $$
  (x_1,y_1) + (x_2,y_2) = (x_1 + x_2,y_1 + y_2)\quad\text{and}\quad
  \alpha(x_1,y_1) = (\alpha x_1, \alpha y_1)
  $$
  for every $(x_1,y_1), (x_2,y_2) \in X\times Y$ and $\alpha\in\bbf$.
  \begin{itemize}
  \item[(a)] Show that $X\times Y$ is a linear space over $\bbf$.
  \item[(b)] Show that
    $$
    \|(x,y)\| = \|x\|_X + \|y\|_Y
    $$
    defines a norm on $X\times Y$.
  \item[(c)] Show that a sequence $\{(x_n,y_n)\}$ of $X\times Y$
    converges to $(x,y)\in X\times Y$ as $n\to\infty$ if and only if
    $\{x_n\}$ converges to $x\in X$ as $n\to\infty$ and $\{y_n\}$
    converges to $y\in Y$ as $n\to\infty$.
  \item[(d)] Show that $\{(x_n,y_n)\}$ is Cauchy  in $X\times
    Y$ if and only if $\{x_n\}$ is Cauchy in $X$ and $\{y_n\}$ is
    Cauchy in $Y$.
  \end{itemize}
\end{Exec}

\begin{proof}
  \begin{itemize}
  \item[(a)] For every $(x_1,y_1) , (x_2, y_2), (x_3, y_3) \in X \times Y$ and
    $\alpha, \beta \in \bbf$,
    since we have
    \begin{romlist}
    \item $(x_1,y_1) + (x_2,y_2) = (x_1+x_2,y_1+y_2) = (x_2+x_1,y_2+y_1) =
      (x_2,y_2) + (x_1,y_1)$ , $(x_1,y_1) + ((x_2,y_2) + (x_2,y_2)) =
      (x_1,y_1) + (x_2+x_3, y_2+y_3) = (x_1+x_2+x_3, y_1+y_2+y_3)
      = ((x_1,y_1) + (x_2,y_2))+(x_3,y_3)$
    \item $(x_1,y_1)+(0,0) = (x_1+0,y_1+0) = (x_1,y_1)$
    \item $(x_1,y_1)+(-x_1,-y_1) = (x_1+(-x_1), y_1+(-y_1) = (0,0)$
    \item $1(x_1,y_1) = (1x_1,1y_1) = (x_1,y_1)$,
      $\alpha(\beta(x_1,y_1)) = \alpha(\beta x_1,\beta y_1)
      = (\alpha\beta x_1, \alpha\beta y_1) = (\alpha\beta)(x_1,y_1) $,
    \item $\alpha((x_1,y_1)+(x_2,y_2) = \alpha (x_1+x_2, y_1+y_2)
      = (\alpha x_1 + \alpha x_2, \alpha y_1 + \alpha y_2)
      = \alpha(x_1,y_1) +\alpha(x_2,y_2)$,
      $(\alpha+\beta)(x_1,y_1) = ((\alpha+\beta)x_1,(\alpha+\beta)y_1)
      = (\alpha x_1+\beta x_1,\alpha y_1+\beta y_1)
      = (\alpha x_1, \alpha y_1) +(\beta x_1, \beta y_1)
      = \alpha (x_1,y_1) + \beta (x_1,y_1)
      $,
    \end{romlist}
    $X \times Y$ is a linear space over $\bbf$.
  \item[(b)] Since
    \begin{romlist}
    \item $\|(x_1,y_1)\| = \|x_1\|_X + \|y_1\|_Y \geq 0$,
      and $\|(x_1,y_1)\| = 0$ if and only if $\|x_1\|=0$ and $\|y_1\|=0$
      so that $(x_1,y_1) = (0,0)$,
    \item $\|\alpha(x_1,y_1)\| = \|(\alpha x_1, \alpha y_1)\|
      = \|\alpha x_1\|_X + \|\alpha y_1\|_Y = \alpha \|x_1\|_X +\alpha \|y_1\|_Y
      = \alpha (\|x_1\|_X + \|y_1\|_Y) = \alpha \|(x_1,y_1)\|$,
    \item $\|(x_1,y_1)+(x_2,y_2)\| = \|(x_1+x_2,y_1+y_2)\|
      = \|x_1+x_2\|_X + \|y_1+y_2\|_Y
      \leq \|x_1\|_X + \|x_2\|_X + \|y_1\|_Y + \|y_2\|_Y
      = \|(x_1,y_1)\| + \|(x_2,y_2)\|$,
    \end{romlist}
    it follows that $\|\cdot\|$ is a norm on $X\times Y$.
  \item[(c)] A sequence $\{(x_n,y_n)\}$ of $X\times Y$ converges to $(x,y)$
    if and only if $\|(x_n,y_n) - (x,y)\|
    = \|x_n-x\| + \|y_n-y\| \rightarrow 0 $ as $n \rightarrow \infty$ so that
    $\|x_n-x\| \rightarrow 0$ and $\|y_n-y\| \rightarrow 0$.
    Hence the sequence $\{(x_n,y_n)\}$
    converges to $(x,y)\in X\times Y$ as $n\to\infty$ if and only if
    $\{x_n\}$ converges to $x\in X$ as $n\to\infty$
    and $\{y_n\}$ converges to $y\in Y$.
  \item[(d)] $\{(x_n,y_n)\}$ is Cauchy if there exists $N>0$
    such that $\|(x_n,y_n) - (x_m,y_m)\| = \|x_n-x_m\| + \|y_n-y_m\|
    \leq \varepsilon$ for every $\varepsilon >0, n,m>N$.
    Hence $\|x_n-x_m\| < \varepsilon $ and $\|y_n-y_m\|<\varepsilon$ so
    that $\{x_n\}$ and $\{y_n\}$ are Cauchy.
    On the other hand, if $\{x_n\}$ and $\{y_n\}$ are Cauchy.
    For every $\varepsilon>0$ there exist integers $M_1,M_2>0$ such that
    $\|x_n-x_m\|< \varepsilon/2$ for every $n,m > M_1$ and
    $\|y_n-y_m\|< \varepsilon/2$ for every $n,m > M_2$.
    Let $N = \max\{M_1,M_2\}$, then $\|(x_n,y_n) - (x_m,y_m)\|
    = \|x_n-x_m\| + \|y_n-y_m\| < \varepsilon$
    so that $\{(x_n,y_n)\}$ is Cauchy.
  \end{itemize}
\end{proof}

\begin{Exec}\label{Exec2.3}
  Consider the real
  linear space $\br^2$.  For every $x = (x_1, x_2)\in\br^2$, let
  $\|x\|_1 = |x_1| + |x_2|$.
  \begin{itemize}
  \item[a)]  Show that$\|\cdot\|_1$ defines a norm on $\br^2$.
  \item[b)]  Sketch the unit circle $\{x \in\br^2 : \|x\|_1 = 1\}$.
  \end{itemize}
\end{Exec}

\begin{proof}
  Since $(\br,|\cdot|)$ is an normed linear space over $\br$,
  $\|\cdot\|_1$ defines a norm on $\br^2$ follows from  Exercise~\ref{Exec2.2}.
  The unit circle $\{x \in\br^2 : \|x\|_1 = 1\}$ looks like follow figure.
  
  \includegraphics{che2.2030}
\end{proof}
\begin{Exec}
  Show that the discrete metric on a
  vector space $X\not= \{0\}$ cannot be obtained from a norm.
\end{Exec}

\begin{solution}
  Suppose that $X$ is a vector space on $\bbf$ and that $d$ is the discrete metric on $X$.
  Since $X \neq \{0\}$, there exists a vector $v$ such that $v\neq 0$.
  Hence $d(2v,0) =1 \neq 2 = 2 d(v,0)$, so that $d$ cannot be obtained from a norm by
  Remark~\ref{Rek2.2.2}.
\end{solution}

\begin{Exec}\label{Exec2.5}
  Let $(X,\|\cdot\|)$ be s normed linear space.
  Define $\tilde{d}$ by
  $$
  \tilde{d}(x,x)= 0,  \qquad \tilde{d}(x,y)=\|x-y\|+1\quad\text{if}\
  x\not= y.
  $$
  Prove that $\tilde{d}$ is a metric on $X$ but not the metric induced
  by the norm $\|\cdot\|$.
\end{Exec}

\begin{proof}
  Clearly $\tilde{d}(x,y) \geq 0$ and $\tilde{d}(x,y) = 0$
  if and only if $x=y$.
  Since $\tilde{d}(x,y)=\|x-y\| +1 = =\|y-x\| +1 = \tilde{d}(y,x)$ for every
  $x\neq y \in X$, $\tilde{d}(x,y) = \tilde{d}(y,x)$ for every $x, y \in X$.
  $\tilde{d}(x,y) \leq \|x-y\| + 1
  \leq \|x-z\| + 1 + \|y-z\| \leq d(x,z)+ d(y,z)$ if $x \neq y\in X$.
  the triangle inequality is clearly true when $x=y$.
  Hence $\tilde{d}$ is a metric on $X$.
  $\tilde{d}$ is not the metric induced by the norm $\|\cdot\|$
  follows from the fact that $\tilde{d}(\alpha x,0)\not=|\alpha|\tilde{d}(x,0)$, in general.
\end{proof}

\begin{Exec}
  Suppose that $(X,\|\cdot\|_1)$ and $(Y,\|\cdot\|_2)$ are Banach spaces.
Let $Z=X \times Y$ with the norm
$$
\|(x,y)\|=\|x\|_1+\|y\|_2.
$$
Prove that $Z$ is a Banach space.
\end{Exec}
\begin{proof}
  By the Exerciese~\ref{Exec2.2}, $(Z, \|\cdot\|)$ is a normed linear space.
  For every Cauchy sequence $\{(x_n,y_n)\}$ in $Z$, $\{x_n\}$ is Cauchy in $X$
  and $\{y_n\}$ is Cauchy in $Y$.
  Since $(X,\|\cdot\|_1)$ and $(Y,\|\cdot\|_2)$ are Banach spaces,
  $\{x_n\}$ and $\{y_n\}$ are convergent. Hence $\{(x_n,y_n)\}$ is
  convergent
  so that $Z$ is a Banach space.
\end{proof}

\begin{Exec}
  Prove that $L^\infty(E)$ of Example~\ref{Exam2.3.5}
  is a Banach space.
\end{Exec}

\begin{proof}
  Suppose that $\{x_i\} \in L^\infty(E)$ such that
  $\sum_{i=1}^{\infty} \|x_i\|_\infty < \infty$.
  By the Example~\ref{Exam2.3.6}, there exists a set $E_i$ such that
  $\|x_i\| = \sup_{E\setminus E_0} |x(t)|$ and $\text{meas}(E_i) = 0$ for every $i \geq 1$.
  Let $E_0 = \cup_{i\geq 1} E_i$, then $\text{meas}(E_0) = 0$.
  So $\|\sum_{i=1}^{\infty} x_i \| \leq \sup_{E\setminus E_0} |\sum_{i=1}^{\infty} x_i(t)|
  \leq \sum_{i=1}^{\infty} \sup_{E\setminus E_0} |x_i(t)|
  \leq \sum_{i=1}^{\infty} \sup_{E\setminus E_i} |x_i(t)|
  = \sum_{i=1}^{\infty} \|x_i\|_\infty < \infty$.
  Hence $L^\infty(E)$ is complete so that it is a Banach space.
\end{proof}

\begin{Exec}\label{Exec2.8}
   Prove that $L^\infty(E)$ is not separable.
\end{Exec}

\begin{proof}
  Consider $B[a,b]$ defined in Exercise~\ref{Exec1.16}.
  Clearly $B[a,b]$ is the subspace of $L^\infty[a,b]$.
  If $L^\infty[a,b]$ is separable,
  we get $B[a,b]$ is separable by Exercise~\ref{Exec1.15},
  a contradiction with Exercise~\ref{Exec1.16}.
\end{proof}

\begin{Exec}
  Prove that $\ell^p$ ($p\geq 1$) is a separable Banach space.
\end{Exec}
\begin{proof}
  Let $E_n = \{(x_1, \ldots, x_n, \ldots):
  x_i \in \bq, x_i = 0 \text{ when } i\geq n \}$ and
  $E = \cup_{n\in\bn} E_n$.
  Clearly $E_n \subset \ell^p$, $E_n$ is countable and so is $E$.
  For every $x= (x_1, \ldots, x_n, \ldots) \in \ell^p$ and
  $\varepsilon > 0$, since $\|x\|^p = \sum_{k=1}^\infty |x_k|^p < \infty$,
  there exists $N >0$ such that
  $\sum_{k=N}^{\infty}|x_k| \leq \varepsilon/2$.
  Let $y = \{y_i\} \in E_{N}$ such that $y_i \in \bq$ and
  $|x_i-y_i| \leq (\varepsilon/2N)^{1/p}$ when $i < N$.
  Hence $d(x,y)^p = \sum_{k=1}^\infty |x_k-y_k|^p = \sum_{k=1}^{N-1}|x_k-y_k|^p
  + \sum_{k=N}{\infty}|x_k|^p < (N-1)(\varepsilon/2N) + \varepsilon/2
  < \varepsilon$ so that there exists a sequence in $E\cap l^p$
  So $\ell^p$ is separable.

  Suppose that $\sum_{j=1}^{n} \|(x_k^{(j)})\| < \infty$.
  By the Minkowski inequality, we get
  $$
  (\sum_{k=1}^{\infty} |\sum_{j=1}^{n} x_k^{(j)}|^p)^{1/p}
  \leq \sum_{j=1}^{n} (\sum_{k=1}^{\infty} |x_k^{(j)}|^p)^{1/p}.
  $$
  Taking $n\rightarrow \infty$ in the above, we have
  $\|\sum_{j=1}^{\infty} (x_k^{(j)})\|
  \leq \sum_{j=1}^{n} \|(x_k^{(j)})\| < \infty$.
  Hence $\ell^p$ is complete.
  Thus $\ell^p$ is a separable Banach space.
\end{proof}

\begin{Exec}
  Prove that the space $c$ and $c_0$ are separable Banach spaces,
  where $c,c_0$ are the subspaces of $\ell^\infty$ in Example~\ref{Exam1.2.9}.
\end{Exec}

\begin{proof}
  Let $E_n = \{\{x_i\}: x_i \in \bq, x_i = x_n \text{ when } i\geq n \}$ and
  $E = \cup_{n\in\bn} E_n$.
  Clearly $E_n \subset c$ and $E_n$ is countable, then $E$ is countable.
  For every $x=\{x_n\} \in c$ and $\varepsilon > 0$, there exists $N >0 $
  such that $|x_n - x_m| \leq \varepsilon/2$ for every $n,m > N$.
  Let $y = \{y_i\} \in E_{N+1}$
  such that $y_i \in \bq$ and $|x_i-y_i| \leq \varepsilon/2$ when $i \leq N+1$.
  Hence $d(x,y) = sup_{i\in\bn} |x_i-y_i|
  < \max\{\varepsilon/2, \sup_{i>N}\{|x_i-x_{N+1}|+|x_{N+1}-y_{i}|\}
  \leq \varepsilon$
  so that $c$ is separable.

  By Exercise~\ref{Exec1.15}, $c_0$ is separable
  follows from that $c_0 \subset c$.

  By Exercise~\ref{Exec1.18} $c, c_0$ is complete.

  Since $d(x,y) = \sup_{i\in\bn} |\eta_i - \xi_i| = d(x-y,0)$
  and
  $$
  d(\alpha x, 0) = \sup_{i\in\bn} |\alpha \eta_i|
  = |\alpha| \sup_{i\in\bn} |\eta_i| = |\alpha| d(x,0)
  $$
  for every $x = \{\eta_i\}, y = \{\xi_i\}$.
  Hence $\|x\| = d(x,0)$ is a norm on $c(or c_0)$ induced by the metric $d$.

  Thus $c, c_0$ are separable Banach spaces.
\end{proof}

\begin{Exec}
  Let $E$ be a Lebesgue \text{meas}urable set of $\br$ with
  \text{meas}$(E)<\infty$. Denote $\|\cdot\|_p$ and $\|\cdot\|_\infty$ by the
  norms of $L^p(E)$ and $L^\infty(E)$ ($p\geq 1$), respectively. Prove
  that $L^\infty(E)\subset L^p(E)$ for all $p\geq 1$ and
  $$
  \lim_{p\to\infty}\|x\|_p=\|x\|_\infty
  $$
  for each $x\in L^\infty(E)$.
\end{Exec}

\begin{proof}
  Since
  $$
  \int_E |f(x)|^p dx \leq \int_E \esssup_E |f(x)|^p dx
  = (\esssup_E|f(x)|)^p \text{meas}(E) < \infty
  $$
  for every $f \in L^\infty(E)$ and $p \geq 1$,
  $L^\infty(E)\subset L^p(E)$.

  Let $M = \esssup_E |f(x)|$.
  Since the difinition of $\esssup$, there exists a \text{meas}urable set
  $A \in E$ such that $A=\{x: f(x) > M'\}$ and $\text{meas}(A) > 0$ for every $M'<M$.
  Consider that $\|f\|_p = (\int_E |f(x)|^p dx)^{1/p}
  \geq (\int_A |f(x)|^p dx)^{1/p} \geq M' (\text{meas}(A))^{1/p}$ for every $p>1$,
  we get $\varliminf_{p\rightarrow\infty} |f(x)|_p \geq M'$.
  Take $M' \rightarrow M$ we get $\varliminf_{p\rightarrow\infty} |f(x)|_p \geq M$.
  On the other hand, $\|f\|_p = (\int_E |f(x)|^p dx)^{1/p} \leq M (\text{meas}(E))^{1/p}$.
  Hence $\varlimsup_{p\rightarrow\infty} |f(x)|_p \leq M$ so that
  $\lim_{p\rightarrow\infty} |f(x)|_p = M$.
\end{proof}

\begin{Exec}
  Theorem~\ref{Thm2.4.3} implies that
  $\|\cdot\|_2$ and $\|\cdot\|_\infty$ in Example~\ref{Exam2.4.1} are
  equivalent.  Give a direct proof of this fact.
\end{Exec}
\begin{proof}
  For every $x=(x_1, \ldots, x_n) \in \bbf^n$, since
  \[
  \|x\|_\infty = \max_{1\leq i\leq n}\{|x_i|\}
  \leq n (\sum_{1\leq i\leq n} |x_i|^2)^{1/2} = n \|x\|_2
  \]
  and
  \[
  \|x\|_2 = (\sum_{1\leq i\leq n} |x_i|^2)^{1/2}
  \leq n^{1/2} \max_{1\leq i\leq n}\{|x_i|\} \leq n \|x\|_\infty,
  \]
  we get $1/n \|x\|_2 \leq \|x\|_\infty \leq n \|x\|_2$.
  Thus $\|\cdot\|_2$ and $\|\cdot\|_\infty$ are equivalent.
\end{proof}

\begin{Exec}
  Let $P[0, 1]$ denote
  the complex vector space of all complex valued polynomials
  defined on $[0, 1]$. This can be viewed as a linear subspace of
  $C[0, 1]$. Show that the two norms
  $$
  \|f\|_\infty =\sup_{t\in [0,1]} |f(t)|\quad\text{and}\quad
  \|f\|_1=\int_0^1 |f(t)| dt
  $$
  are not equivalent on $P[0, 1]$.
\end{Exec}
\begin{proof}
  Consider $f_n(x)=x^n \in P[0,1]$,
  since $\|f_n\|_\infty = \sup_{t\in [0,1]} |f_n(t)| = 1$
  and $\|f_n\|_1=\int_0^1 |f_n(t)| dt = 1/(n+1)$,
  $\|f_n\|_\infty/\|f_n\|_1 = n+1 \rightarrow \infty$ as $n\rightarrow \infty$.
  If $\|f\|_\infty$ and $\|f\|_1$ are equivalent, there exists $K>0$ such that
  $\|f_n\|_\infty/\|f_n\|_1 \leq K$, a contradiction.
  Hence  $\|f\|_\infty$ and $\|f\|_1$ are not equivalent.
\end{proof}

\begin{Exec}
  Show that all complex $m\times n$
  matrices $A=(\alpha_{jk})$ with fixed $m$ and $n$ constitute an
  $mn$-dimensional linear space $Z$.  Show that all norms on $Z$ are
  equivalent.  What would be the analogues of $\|\cdot\|_1$,
  $\|\cdot\|_2$ and $\|\cdot\|_\infty$ in Example~\ref{Exam2.4.1} for
  the present space $Z$?
\end{Exec}
\begin{proof}
  By Theorem~\ref{Thm2.4.3}, all norms on $Z$ are equivalent.
  We get
  \[
  \|x\|_1 = \sum_{j=1}^{m} \sum_{k=1}^{n} |a_{jk}|,
  \]
  \[
  \|x\|_2 = (\sum_{j=1}^{m} \sum_{k=1}^{n} |a_{jk}|^2)^{1/2}
  \]
  and
  \[\|x\|_\infty = \max_{\substack{1\leq j\leq m\\ 1\leq k\leq n}} |a_{jk}|\].
\end{proof}

\begin{Exec}
  Let $S$ and $\{x_n\}$ be in Example~\ref{Exam2.5.2}.
  Prove that $x_n\in S$ for all $n\in\bn$ and draw sketch maps
  of $x_n$ for $n=1,2,3,4,5$.
\end{Exec}

\begin{proof}
  Since $\|x_n\| = \left(\int_0^1 |x_n(t)|^2dt|\right)^{1/2} = 1-2/3n < 1$
  for ervery $n$, $x_n \in X$. By the definition of $x_n$,
  $x_n(0) = 0$ so that $x_n \in S$.

  \includegraphics{che2.2150}
\end{proof}

\begin{Exec}\label{Exec2.16}
  If $S=\{\{x_n\}\in\ell^2:\ \text{there exists $N\in\bn$ such that
    $x_n=0$ for $n\geq N$}\} $, so that $S$ is linear subspace of
  $\ell^2$ consisting of sequences having only finitely many non-zero
  terms.  Show that $S$ is not closed.
\end{Exec}

\begin{proof}
  Consider that $\xi_n=\{x_i\}$ such that $x_i = i^{-4}$ for $i\leq n$ and
  $x_i =0$ for $i>n$.
  Clearly $\xi_n \in S$ and $\xi_n \rightarrow \{i^{-4}\} \in \ell^2$
  as $n \rightarrow \infty$.
  Since $\{i^{-4}\}\not\in S$, $S$ is not closed.
\end{proof}

\begin{Exec}
  Consider the complex vector space $\ell^\infty$ of all bounded
  infinite sequences $x = (x_1,x_2,x_3\cdots)$ of
  complex numbers. Let $A = \{e_j : j\in N\}$, where
  $$
  e_j = (\underbrace{0,\cdots, 0}_{j-1},1, 0, 0, 0,\cdots)
  \quad\text{for every}\ j\in\bn.
  $$
  Show that $\overline{\text{\rm span}(A)} = c_0$, where $c_0$ is
  in Exercise~\ref{Exec2.8}.
\end{Exec}

\begin{proof}
  Since $A \subset c_0$ and $c_0$ is a linear space,
  $\sspan(A) \subset c_0$.
  So $\overline{\sspan(A)} \subset \overline{c_0} = c_0$
  follows from that $c_0$ is closed.

  For every $x=\{\xi_i\} \in c_0$,
  let $x_n = \sum_{i=0}^{n} \xi_i e_i \in \sspan(A)$.
  We denote that $x_n\rightarrow x$ as $n \rightarrow \infty$.
  In fact, for every $\varepsilon >0$, there exists a integer $N>0$
  such that $|\xi_i| < \varepsilon/2 $ for every $i>N$.
  Hence $\|x_n - x\| = \sup_{i>n} |\xi_i| < \varepsilon$ so that
  $x_n\rightarrow x$ as $n \rightarrow \infty$.
  We get $\overline{\text{\rm span}(A)} \supset c_0$.

  Thus $\overline{\text{\rm span}(A)} = c_0$.
\end{proof}

\begin{Exec}\label{Exec2.18}
  Let $X$ be a normed linear space, let $x\in X\backslash\{0\}$
  and let $Y$ be a linear subspace of $X$.
  \begin{itemize}
  \item[(a)] If there exists $\eta>0$ such that $\{y\in X:\|y\|<\eta\}\subset
    Y$, show that $\dfrac{\eta x}{2\|x\|}\in Y$.
  \item[(b)] If $Y$ is open, show that $Y=X$.
  \end{itemize}
\end{Exec}
\begin{proof}
  \begin{itemize}
  \item[(a)] Since $\|\eta x/2\|x\|\| = \eta/2 <\eta$, it follows that
    $\eta x/2\|x\|\in \{y\in X:\|y\|<\eta\}\subset Y$.
  \item[(b)] Clearly that $0 \in Y$. By the precondition,
    there exists a real number $\eta>0$ such that
    $\{y\in X:\|y-0\|<\eta\} \subset Y$.
    By Exercise~\ref{Exec2.18}(a), for every $x \in X\setminus\{0\}$,
    $\dfrac{\eta x}{2\|x\|}\in Y$.
    Since $Y$ is a linear space,
    $x \in \sspan\{\eta x/2\|x\|\} \subset \sspan Y = Y$ so that
    $X \subset Y$.
    Take $Y \subset X$, we get $Y=X$.
  \end{itemize}
\end{proof}

\begin{Exec}
  Let $X$ be a normed linear space, let $T=\{x\in X:\|x\|\leq
  1\}$ and let $S=\{x\in X: \|x\|<1\}$.
  \begin{itemize}
  \item[(a)] Show that $T$ is closed.
  \item[(b)] If $x\in T$ and $x_n=(1-1/n)x$ for $n\in\bn$, show
    that $x_n\to x$ as $n\to\infty$ and hence show that
    $\overline{S}=1$.
  \end{itemize}
\end{Exec}

\begin{proof}
  \begin{itemize}
  \item[(a)] Since $\|\cdot\|$ is a continuous mapping on $X$
    and $[0,1]$ is closed,
    $T=\{x\in X:\|x\|\leq 1\} = \{x\in X:\|x\|\in [0,1]\}$ is closed.
  \item[(b)] Since $\|x_n -x\| = \|-x/n\| \leq 1/n \rightarrow 0$
    as $n \rightarrow \infty$, it follows that $x_n\rightarrow x$.
    Take $\|x_n\| = \|(1-1/n) x\| \leq (1-1/n) < 1$, we get$x_n \in S$.
    Hence $T\subset \overline{S}$ so that $\overline{S}=T$.
  \end{itemize}
\end{proof}

\begin{Exec}
  Let $X$ be a normed linear space with $X\not=\{0\}$.
  Show that $X$ is a Banach space if and only if the set $S=\{x\in
  X:\|x\|=1\}$ is complete in $X$.
\end{Exec}
\begin{proof}
  Suppose that $S$ is complete. Let $\{x_n\}$ is a Cauchy sequence in $X$.
  Since $|\|x_n\|-\|x_m\|| < \|x_n - x_m\|$ for every $n,m\in \bn$,
  $\{\|x_n\|\}$ is a Cauchy sequence in $\br$
  so that $\{\|x_n\|\}$ is convergence.

  If $\|x_n\|\rightarrow 0$,
  $x_n \rightarrow 0$ by the continuous of $\|\cdot\|$.

  If $\|x_n\|\rightarrow c$ such that $c>0$, there exists a number $N_1>0$
  such that $c/2 < \|x_n\| < 3c/2$ for every $n>N_1$.

  Since $\{x_n\}$ is a Cauchy sequence , there exists a number $N>N_1$
  such that $\|x_n-x_m\|<\varepsilon$ for $\varepsilon >0$.
  Take
  \[
  \begin{split}
    &\left\|\frac{x_n}{\|x_n\|} - \frac{x_m}{\|x_m\|}\right\| \\
    = & \left\|\frac{\|x_m\| x_n - \|x_n\| x_m}{\|x_n\|\|x_m\|}\right\| \\
    = & \frac{4}{c^2 }
    \left\| (x_n-x_m) \|x_m\| + x_m (\|x_m\|-\|x_n\|) \right\| \\
    \leq&  \frac{4}{c^2} \left(
      \|x_n-x_m\| \|x_m\| + \|x_m\| | \|x_m\| - \|x_n\|  |
    \right) \\
    \leq&  \frac{4}{c^2}\frac{3c}{2} \varepsilon  = \frac{6}{c} \varepsilon,
  \end{split}
  \]
  we get $\{x_n/\|x_n\|\}$ is a Cauchy sequence in $S$.
  Since $S$ is complete, there exists a point $x$ in $S$ such that
  $x_n/\|x_n\|\rightarrow  x$ so that $x_n \rightarrow c\,x$.
  Hence $X$ is complete.

  On the other hand, if $X$ is complete,
  $S$ is complete follows from that $S$ is closed in $X$.
\end{proof}
\docend
